We know that a higher-order ode can be converted to dynamical system by replacing each higher-order derivative by a new variable. What about inverse problem? Does a dynamical system convert to a higher-order ode?
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$\begingroup$ A dynamical system is a system of ODE. Are you asking whether a system of ODE can be converted to a single higher order ODE? $\endgroup$– Alexandre EremenkoCommented Aug 29, 2013 at 5:05
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$\begingroup$ Yes, that is what I mean. $\endgroup$– ShuchangCommented Aug 29, 2013 at 5:06
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$\begingroup$ Then you have to specify that you mean by "equivalent". $\endgroup$– Alexandre EremenkoCommented Aug 29, 2013 at 13:13
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$\begingroup$ I mean If I know the solution of one side, so do of the other through some one-one-correspondence mapping. $\endgroup$– ShuchangCommented Aug 29, 2013 at 13:27
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For example, consider a system such as $$ \dot{x} = x, \dot{y} = y $$ for which the origin is a proper node. There is no autonomous second-order ODE $\ddot{z} = f(z, \dot{z})$ that has a proper node: if the linearization at an equilibrium point has a double eigenvalue, it is an improper node. So there can be no smooth conversion of the system to an autonomous second-order ODE.
answered Aug 29, 2013 at 7:04